Simplify the following expression: $\dfrac{99z^5}{72z^4}$ You can assume $z \neq 0$.
Solution: $ \dfrac{99z^5}{72z^4} = \dfrac{99}{72} \cdot \dfrac{z^5}{z^4} $ To simplify $\frac{99}{72}$ , find the greatest common factor (GCD) of $99$ and $72$ $99 = 3 \cdot 3 \cdot 11$ $72 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$ $ \mbox{GCD}(99, 72) = 3 \cdot 3 = 9 $ $ \dfrac{99}{72} \cdot \dfrac{z^5}{z^4} = \dfrac{9 \cdot 11}{9 \cdot 8} \cdot \dfrac{z^5}{z^4} $ $\phantom{ \dfrac{99}{72} \cdot \dfrac{5}{4}} = \dfrac{11}{8} \cdot \dfrac{z^5}{z^4} $ $ \dfrac{z^5}{z^4} = \dfrac{z \cdot z \cdot z \cdot z \cdot z}{z \cdot z \cdot z \cdot z} = z $ $ \dfrac{11}{8} \cdot z = \dfrac{11z}{8} $